Basic Open Sets

Let be an open subset of Spec A. Suppose so that . Pick an . Since we have . It remains to show . Indeed if then certainly so we have . ♦

With this we have:

Proposition 2.

is quasi-compact, in the sense that every open cover has a finite subcover.

Note

The modern topological term for this is just compact. However, the term quasi-compact is firmly entrenched in the literature for commutative algebra and algebraic geometry so we will respect the tradition here.

Proof

To prove quasi-compactness, it suffices to check covering via basic open sets. Thus suppose is a basic open covering; we need to show there is a finite subcover. Now where , i.e. the ideal generated by the . This can only happen if so we can write

for finitely many choices of . Hence so we have

♦

More on Zariski Topology

Let for a ring A. Immediately, we see some differences between X and affine varieties. Firstly, points of X are prime ideals and often is not closed in X. More generally, we have the following.

Proposition 3.

For , the closure of is . In particular, closed if and only if is a maximal ideal, in which case we call it a closed point.

Proof

The closure C of is the intersection of all containing . This includes so we have . On the other hand, any closed subset containing gives so . Thus . ♦

For example, if A is a domain, then (0) is a prime ideal so is the closure of the point (0). In a topological space X, a generic point of X is a such that the closure of is X.

Note

In topological lingo, this says that Spec A is not a T1 space. It is, however, a T0 space. Compare this with the case of affine varieties which are T1 (i.e. all points are closed), but not T2. [For example, has the cofinite topology.]

Elements of Spec ℤ

Nilradical and Jacobson Radical

Definition.

The intersection of all prime ideals of A is called the nilradical of A, denoted . From the previous article,

Correspondingly, the intersection of all maximal ideals of A is called the Jacobson radical of A, denoted by .

We have the following classification for Jacobson radical.

Proposition 4.

We have .

Proof

(⊇) If for maximal ideal then since we have so there exists such that so is not a unit, i.e. .

(⊆) Suppose ; let . If (1 + xy) is not a unit, it is contained in some maximal ideal . Then , or else we would have . ♦

Philosophy

Following affine varieties, we now look at rings from a new point of view.

“Points” correspond to prime ideals .

“Functions on points” correspond to elements .

“Function vanishes on point” corresponds to .

Thus the nilradical (resp. Jacobson radical) is the set of all functions vanishing on all points (resp. closed points). In a way, non-closed points act as “irreducible closed sets” of closed points, so one naturally asks when the nilradical is equal to the Jacobson radical. When this holds, the ring in question is called a Jacobson ring.

Connected Components

Proposition 5.

Let be a product of rings. Then is the topological disjoint union of and .

Conversely suppose where V, W are disjoint closed subsets, then

.

Proof

The projection maps gives continuous map which takes . This map is injective, continuous and takes closed set to closed set . Hence it is a subspace embedding.

Similarly, we have a subspace embedding ; by exercise A here, the two images form a disjoint union so the first claim is done.

For the second claim, write and so

Thus ; pick , such that so (x) and (y) are coprime ideals. Now which is contained in the nilradical since it is contained in all prime ideals. Thus for some n > 0 so are coprime ideals with product zero. By Chinese Remainder Theorem this gives .

Since we have ; similarly . Since and also form a disjoint union of Spec A by the first part, we are done. ♦

Pictorially we have

Irreducible Spaces

Proposition 6.

The space is irreducible if and only if the nilradical is a prime ideal.

Proof

(⇐) If the nilradical is prime , and Spec A is a union of two closed subsets, one of them must contain ; hence it must contain all primes. Thus Spec A is irreducible.

(⇒) Conversely, suppose Spec A is irreducible. Pick with . Then

By irreducibility, either V(x) or V(y) is the whole Spec A. Assume the former; then x lies in all prime ideals of A so it is nilpotent, i.e. . ♦

In particular, we have:

Corollary 1.

If is irreducible then it has a generic point.

Prime Chains

Finally, another important aspect of Spec A are the prime chains.

Definition.

A prime chain of A is a strictly increasing sequence of prime ideals

of A; the length of the chain is d.

For example, in the ring for a field X, we have which is a prime chain of length 2.

Definition.

The Krull dimension (or just dimension) of a ring A is the supremum of all lengths of prime chains in A.

Examples

1. The dimension of a field is 0.

2. The dimension of ℤ is 1.

3. We have , since for any prime chain of A, we have the prime chain of A[X]:

Note that these ideals are all prime since and .

4. From the above, we have and . Eventually, we will see that equality holds!